240 v incandescent light bulb
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Hameedulla-Ekhlas
Senior Member
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see this equation P = V sq / R
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- New Jersey
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If by brilliance you mean the amount of light, it will much less than half.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Paul, welcome to the forum! 
Simply, since power is amps x volts, when the voltage is halved, so is the resultant current.
1/2 x 1/2 = 1/4
(This is presuming a constant impedance.)
Simply, since power is amps x volts, when the voltage is halved, so is the resultant current.
1/2 x 1/2 = 1/4
(This is presuming a constant impedance.)
More than 15W and probably much more. The resistance of the filament varies very significantly with temperature with the cold resistance being a fraction of the hot resistance.
quogueelectric
Senior Member
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- new york
Then if the hot resistance were higher the wattage would be less.............much less. I am going with the 15w answer...............
Perhaps I didn't explain it very well.
For the answer to be 15W, it would need to be the same resistance at 120V as it is at 240V.
But, at rated voltage and power output the filament would be significantly hotter and thus significantly higher resistance than at half voltage. The lower resistance at 120V means that the dissipated power will be more than 15W.
broadgage
Senior Member
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- London, England
In practice the wattage will be a lot more than 15 watts, about 22 watts in practice.
For a constant resistance, the calculations above are correct, half the voltage would result in half the current, which equals one qaurter of the wattage.
However the resistance of an incandesent lamp is not constant, it declines markedly at a lower operating temperatures, and the wattage will be appreciably more than might be expected.
Why not try the experiment ?
Connect a 120 volt 60 watt lamp to a 120 volt supply and measure the current, it should be about 0.5 amp.
Now connect two such lamps in series such that each only receives 60 volts, and again measure the current, it will be less than 0.5 amp, but not as low as the 0.25 amps that might be expected.
Safety note, any experienced electrician may easily do this simple experiment safely, any non-electrician reading this should think very carefuly before experimenting with line voltage.
For a constant resistance, the calculations above are correct, half the voltage would result in half the current, which equals one qaurter of the wattage.
However the resistance of an incandesent lamp is not constant, it declines markedly at a lower operating temperatures, and the wattage will be appreciably more than might be expected.
Why not try the experiment ?
Connect a 120 volt 60 watt lamp to a 120 volt supply and measure the current, it should be about 0.5 amp.
Now connect two such lamps in series such that each only receives 60 volts, and again measure the current, it will be less than 0.5 amp, but not as low as the 0.25 amps that might be expected.
Safety note, any experienced electrician may easily do this simple experiment safely, any non-electrician reading this should think very carefuly before experimenting with line voltage.
gar
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- EE
100314-0733 EST
For a tungsten filament lamp the wattage is about 1/3 at half voltage, and intensity is about 15%.
See my plot P9 at my website http://beta-a2.com/EE-photos.html
.
For a tungsten filament lamp the wattage is about 1/3 at half voltage, and intensity is about 15%.
See my plot P9 at my website http://beta-a2.com/EE-photos.html
.
dbuckley
Senior Member
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- Canterbury, New Zealand
Heres an actual example I did a couple of years ago using a 240V 500W linear halaogen lamp.
Voltage and Amps measured, resistance and wattage calculated, supply is genuine sine AC through a variac. As can be seen the resistance of the filament is anything but constant...
Voltage and Amps measured, resistance and wattage calculated, supply is genuine sine AC through a variac. As can be seen the resistance of the filament is anything but constant...
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
If I recall correctly, the approximate equations are:
current is proportional to the square root of voltage
(meaning that if you multiply the voltage to 2x, the current goes up to 1.414x)
since power is the product of voltage and current, this means that power scales as the 1.5x power
(I checked a couple of points in the data that dbuckley provided, and got rough agreement)
The same set of equations said that visible light output scaled as the square of the power. As the light gets dimmer it gets much less efficient. This scaling does not seem to agree with the plots that gar posted, but this is almost certainly the difference between theory and practise, unless the photocell that gar was using is more sensitive to red and IR than the human eye. As the light gets dimmer, the spectrum shifts toward the red.
-Jon
current is proportional to the square root of voltage
(meaning that if you multiply the voltage to 2x, the current goes up to 1.414x)
since power is the product of voltage and current, this means that power scales as the 1.5x power
(I checked a couple of points in the data that dbuckley provided, and got rough agreement)
The same set of equations said that visible light output scaled as the square of the power. As the light gets dimmer it gets much less efficient. This scaling does not seem to agree with the plots that gar posted, but this is almost certainly the difference between theory and practise, unless the photocell that gar was using is more sensitive to red and IR than the human eye. As the light gets dimmer, the spectrum shifts toward the red.
-Jon
gar
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- Ann Arbor, Michigan
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- EE
100315-0613 EST
winnie:
The photocell used was Clairex 505L. I do not know where any datasheets are at present. A Google search seemed to imply the response was similar to the human eye. "spectral response of clairex 605L" . However, I could not access the raw data of the Sage Journal, they want $. Similar problem on other sources that might have curves.
Found this reference for the 505L http://smartech.gatech.edu/bitstream/1853/1034/1/3269_001_01121976.pdf Figure 8, doing a find on 505L brings one to the figure quickly.
.
winnie:
The photocell used was Clairex 505L. I do not know where any datasheets are at present. A Google search seemed to imply the response was similar to the human eye. "spectral response of clairex 605L" . However, I could not access the raw data of the Sage Journal, they want $. Similar problem on other sources that might have curves.
Found this reference for the 505L http://smartech.gatech.edu/bitstream/1853/1034/1/3269_001_01121976.pdf Figure 8, doing a find on 505L brings one to the figure quickly.
.
So is it 120x120 dived by 240? I would have got this answer wrong on a test. I thought the watts would stay the same, and the amps would change.
gar
Senior Member
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- Ann Arbor, Michigan
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- EE
100415-0837 EST
zappy:
As you change the voltage to a tungsten filament lamp the filament temperature changes and therefore its resistance changes. You need to know the resistance at the new voltage to determine the current and wattage at that voltage.
.
zappy:
As you change the voltage to a tungsten filament lamp the filament temperature changes and therefore its resistance changes. You need to know the resistance at the new voltage to determine the current and wattage at that voltage.
.
winnie
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- Springfield, MA, USA
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- Electric motor research
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